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Is there a Pandas function to highlight a week's 10 lowest values in a time series?

Rookie here so please excuse my question format:
I got an event time series dataset for two months (columns for "date/time" and "# of events", each row representing an hour).
I would like to highlight the 10 hours with the lowest numbers of events for each week. Is there a specific Pandas function for that? Thanks!

Let's say you have a dataframe df with column col as well as a datetime column.
You can simply sort the column with
import pandas as pd
df = pd.DataFrame({'col' : [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15],
'datetime' : ['2019-01-01 00:00:00','2015-02-01 00:00:00','2015-03-01 00:00:00','2015-04-01 00:00:00',
'2018-05-01 00:00:00','2016-06-01 00:00:00','2017-07-01 00:00:00','2013-08-01 00:00:00',
'2015-09-01 00:00:00','2015-10-01 00:00:00','2015-11-01 00:00:00','2015-12-01 00:00:00',
'2014-01-01 00:00:00','2020-01-01 00:00:00','2014-01-01 00:00:00']})
df = df.sort_values('col')
df = df.iloc[0:10,:]
df
Output:
col datetime
0 1 2019-01-01 00:00:00
1 2 2015-02-01 00:00:00
2 3 2015-03-01 00:00:00
3 4 2015-04-01 00:00:00
4 5 2018-05-01 00:00:00
5 6 2016-06-01 00:00:00
6 7 2017-07-01 00:00:00
7 8 2013-08-01 00:00:00
8 9 2015-09-01 00:00:00
9 10 2015-10-01 00:00:00

I know there's a function called nlargest. I guess there should be an nsmallest counterpart. pandas.DataFrame.nsmallest
df.nsmallest(n=10, columns=['col'])
My bad, so your DateTimeIndex is a Hourly sampling. And you need the hour(s) with least events weekly.
...
Date n_events
2020-06-06 08:00:00 3
2020-06-06 09:00:00 3
2020-06-06 10:00:00 2
...
Well I'd start by converting each hour into columns.
1. Create an Hour column that holds the hour of the day.
df['hour'] = df['date'].hour
Pivot the hour values into columns having values as n_events.
So you'll then have 1 datetime index, 24 hour columns, with values denoting #events. pandas.DataFrame.pivot_table
...
Date hour0 ... hour8 hour9 hour10 ... hour24
2020-06-06 0 3 3 2 0
...
Then you can resample it to weekly level aggregate using sum.
df.resample('w').sum()
The last part is a bit tricky to do on the dataframe. But fairly simple if you just need the output.
for row in df.itertuples():
print(sorted(row[1:]))

How do I create a rolling monthly datetime index for pandas?

I have a DatetimeIndex in pandas and I want to convert it to a rolling DatetimeIndex using the last date in the series.
So if I create a sample datetime index:
dates=pd.DatetimeIndex(pd.date_range(dt(2017,10,1),dt(2018,02,02)))
An example
Input: DatetimeIndex with all dates in the above range:
dates
2017-10-01
2017-10-02
.
.
2018-02-01
2018-02-02
Desired Output: DatetimeIndex with only the 2nd of every month (as that is the last date in the input):
dates
2017-10-02
2017-11-02
2017-12-02
2018-01-02
2018-02-02
Attempts
I've tried
dates[::-1][::30]
and also
dates[dates.apply(lambda x: x.date().day==2)]
Unfortunately months can differ by 30 or 31 days so the first way doesn't work and while the second method works for days in range 1-30, for the 31st it skips every other month. So, for example, if I had:
dates
2017-10-01
2017-10-02
.
.
2018-01-31
I would want:
dates
2017-10-31
2017-11-30
2017-12-31
2018-01-31
while the second method skips November as it doesn't have a 30th.
Is there any way to use RelativeDelta to do this?

You can use the .is_month_end functionality in Pandas. This gives an array of boolean values – True if the date is a month-end, false if otherwise.
import pandas as pd
import datetime as dt
dates=pd.Series(pd.date_range('2017-10-1','2017-12-31'))
print(dates[dates.is_month_end])
Output
DatetimeIndex(['2017-10-31', '2017-11-30', '2017-12-31'], dtype='datetime64[ns]', freq=None)
This will help you filter things.

Add months to a date in Pandas

I'm trying to figure out how to add 3 months to a date in a Pandas dataframe, while keeping it in the date format, so I can use it to lookup a range.
This is what I've tried:
#create dataframe
df = pd.DataFrame([pd.Timestamp('20161011'),
pd.Timestamp('20161101') ], columns=['date'])
#create a future month period
plus_month_period = 3
#calculate date + future period
df['future_date'] = plus_month_period.astype("timedelta64[M]")
However, I get the following error:
AttributeError: 'int' object has no attribute 'astype'

You could use pd.DateOffset
In [1756]: df.date + pd.DateOffset(months=plus_month_period)
Out[1756]:
0 2017-01-11
1 2017-02-01
Name: date, dtype: datetime64[ns]
Details
In [1757]: df
Out[1757]:
date
0 2016-10-11
1 2016-11-01
In [1758]: plus_month_period
Out[1758]: 3

Suppose you have a dataframe of the following format, where you have to add integer months to a date column.
Start_Date
Months_to_add
2014-06-01
23
2014-06-01
4
2000-10-01
10
2016-07-01
3
2017-12-01
90
2019-01-01
2
In such a scenario, using Zero's code or mattblack's code won't be useful. You have to use lambda function over the rows where the function takes 2 arguments -
A date to which months need to be added to
A month value in integer format
You can use the following function:
# Importing required modules
from dateutil.relativedelta import relativedelta
# Defining the function
def add_months(start_date, delta_period):
end_date = start_date + relativedelta(months=delta_period)
return end_date
After this you can use the following code snippet to add months to the Start_Date column. Use progress_apply functionality of Pandas. Refer to this Stackoverflow answer on progress_apply : Progress indicator during pandas operations.
from tqdm import tqdm
tqdm.pandas()
df["End_Date"] = df.progress_apply(lambda row: add_months(row["Start_Date"], row["Months_to_add"]), axis = 1)
Here's the full code form dataset creation, for your reference:
import pandas as pd
from dateutil.relativedelta import relativedelta
from tqdm import tqdm
tqdm.pandas()
# Initilize a new dataframe
df = pd.DataFrame()
# Add Start Date column
df["Start_Date"] = ['2014-06-01T00:00:00.000000000',
'2014-06-01T00:00:00.000000000',
'2000-10-01T00:00:00.000000000',
'2016-07-01T00:00:00.000000000',
'2017-12-01T00:00:00.000000000',
'2019-01-01T00:00:00.000000000']
# To convert the date column to a datetime format
df["Start_Date"] = pd.to_datetime(df["Start_Date"])
# Add months column
df["Months_to_add"] = [23, 4, 10, 3, 90, 2]
# Defining the Add Months function
def add_months(start_date, delta_period):
end_date = start_date + relativedelta(months=delta_period)
return end_date
# Apply function on the dataframe using lambda operation.
df["End_Date"] = df.progress_apply(lambda row: add_months(row["Start_Date"], row["Months_to_add"]), axis = 1)
You will have the final output dataframe as follows.
Start_Date
Months_to_add
End_Date
2014-06-01
23
2016-05-01
2014-06-01
4
2014-10-01
2000-10-01
10
2001-08-01
2016-07-01
3
2016-10-01
2017-12-01
90
2025-06-01
2019-01-01
2
2019-03-01
Please add to comments if there are any issues with the above code.
All the best!

I believe that the simplest and most efficient (faster) way to solve this is to transform the date to monthly periods with to_period(M), add the result with the values of the Months_to_add column and then retrieve the data as datetime with the .dt.to_timestamp() command.
Using the sample data created by #Aruparna Maity
Start_Date
Months_to_add
2014-06-01
23
2014-06-20
4
2000-10-01
10
2016-07-05
3
2017-12-15
90
2019-01-01
2
df['End_Date'] = ((df['Start_Date'].dt.to_period('M')) + df['Months_to_add']).dt.to_timestamp()
df.head(6)
#output
Start_Date Months_to_add End_Date
0 2014-06-01 23 2016-05-01
1 2014-06-20 4 2014-10-01
2 2000-10-01 10 2001-08-01
3 2016-07-05 3 2016-10-01
4 2017-12-15 90 2025-06-01
5 2019-01-01 2 2019-03-01
If the exact day is needed, just repeat the process, but changing the periods to days
df['End_Date'] = ((df['End_Date'].dt.to_period('D')) + df['Start_Date'].dt.day -1).dt.to_timestamp()
#output:
Start_Date Months_to_add End_Date
0 2014-06-01 23 2016-05-01
1 2014-06-20 4 2014-10-20
2 2000-10-01 10 2001-08-01
3 2016-07-05 3 2016-10-05
4 2017-12-15 90 2025-06-15
5 2019-01-01 2 2019-03-01

Another way using numpy timedelta64
df['date'] + np.timedelta64(plus_month_period, 'M')
0 2017-01-10 07:27:18
1 2017-01-31 07:27:18
Name: date, dtype: datetime64[ns]

pick month start and end data in python

I have stock data downloaded from yahoo finance. I want to pickup data in the row corresponding to monthly start and month end. I am trying to do it with python pandas data frame. But I am not getting correct method to get the starting & ending of the month. will be great full if somebody can help me in solving this.
Please note that if 1st of the month is holiday and there is no data for that, I need to pick up 2nd day's data. Same rule applies to last of the month also. Thanks in advance.
Example data is
2016-01-05,222.80,222.80,217.00,217.75,15074800,217.75
2016-01-04,226.95,226.95,220.05,220.70,14092000,220.70
2015-12-31,225.95,226.55,224.00,224.45,11558300,224.45
2015-12-30,229.00,229.70,224.85,225.80,11702800,225.80
2015-12-29,228.85,229.95,227.50,228.20,7263200,228.20
2015-12-28,229.05,229.95,228.00,228.90,8756800,228.90
........
........
2015-12-04,240.00,242.15,238.05,241.10,11115100,241.10
2015-12-03,244.15,244.50,240.40,241.10,7155600,241.10
2015-12-02,250.55,250.65,243.75,244.60,10881700,244.60
2015-11-30,249.65,253.00,245.00,250.20,12865400,250.20
2015-11-27,243.00,250.50,242.80,249.70,15149900,249.70
2015-11-26,241.95,244.90,241.00,242.50,13629800,242.50

First, you should convert your date column to datetime format, then group by month, then sort groupby Series by date and take the first/last from it using head/tail methods, like so:
In [37]: df
Out[37]:
0 1 2 3 4 5 6
0 2016-01-05 222.80 222.80 217.00 217.75 15074800 217.75
1 2016-01-04 226.95 226.95 220.05 220.70 14092000 220.70
2 2015-12-31 225.95 226.55 224.00 224.45 11558300 224.45
3 2015-12-30 229.00 229.70 224.85 225.80 11702800 225.80
4 2015-12-29 228.85 229.95 227.50 228.20 7263200 228.20
5 2015-12-28 229.05 229.95 228.00 228.90 8756800 228.90
In [25]: import datetime
In [29]: df[0] = df[0].apply(lambda x: datetime.datetime.strptime(x, '%Y-%m-%d')
)
In [36]: df.groupby(df[0].apply(lambda x: x.month)).apply(lambda x: x.sort_value
s(0).head(1))
Out[36]:
0 1 2 3 4 5 6
0
1 1 2016-01-04 226.95 226.95 220.05 220.7 14092000 220.7
12 5 2015-12-28 229.05 229.95 228.00 228.9 8756800 228.9
In [38]: df.groupby(df[0].apply(lambda x: x.month)).apply(lambda x: x.sort_value
s(0).tail(1))
Out[38]:
0 1 2 3 4 5 6
0
1 0 2016-01-05 222.80 222.80 217.0 217.75 15074800 217.75
12 2 2015-12-31 225.95 226.55 224.0 224.45 11558300 224.45
You can merge the result dataframes, using pd.concat()

For the first / last day of each month, you can use .resample() with 'BMS' and 'BM' for Business Month (Start) like so (using pandas 0.18 syntax):
df.resample('BMS').first()
df.resample('BM').last()
This assumes that your data have a DateTimeIndex as usual when downloaded from yahoo using pandas_datareader:
from datetime import datetime
from pandas_datareader.data import DataReader
df = DataReader('FB', 'yahoo', datetime(2015, 1, 1), datetime(2015, 3, 31))['Open']
df.head()
Date
2015-01-02 78.580002
2015-01-05 77.980003
2015-01-06 77.230003
2015-01-07 76.760002
2015-01-08 76.739998
Name: Open, dtype: float64
df.tail()
Date
2015-03-25 85.500000
2015-03-26 82.720001
2015-03-27 83.379997
2015-03-30 83.809998
2015-03-31 82.900002
Name: Open, dtype: float64
do:
df.resample('BMS').first()
Date
2015-01-01 78.580002
2015-02-02 76.110001
2015-03-02 79.000000
Freq: BMS, Name: Open, dtype: float64
and
df.resample('BM').last()
to get:
Date
2015-01-30 78.000000
2015-02-27 80.680000
2015-03-31 82.900002
Freq: BM, Name: Open, dtype: float64

Assuming you have downloaded data from Yahoo:
> import pandas.io.data as web
> import datetime
> start = datetime.datetime(2016,1,1)
> end = datetime.datetime(2016,5,1)
> df = web.DataReader("AAPL", "yahoo", start, end)
You simply pick the month end and start rows with:
df[df.index.is_month_end]
df[df.index.is_month_start]
If you want to access a specific row, like the first row of the first starting day of the selected starting days, you simply do:
df[df.index.is_month_start].ix[0]

Resample with custom month-end frequency

I'm looking for an equivalent specification to W-MON (weekly, ending Monday) for monthly data.
Specifically, I have a pandas data frame of daily data, and I want to only take monthly observations, starting with the most recent date and going back monthly.
So if today is 17/06/2016, my date index would be 17/06/2016, 17/05/2016, 17/04/2016... etc.
Right now I can only find month-start and month-end as specifications for df.asfreq().
Thanks.

You can create the relevant dates using relativedelta and select using .loc[]:
from datetime import datetime
from dateutil.relativedelta import relativedelta
from pandas_datareader.data import DataReader
Using daily sample data:
stock_data = DataReader('FB', 'yahoo', datetime(2013, 1, 1), datetime.today()).resample('D').fillna(method='ffill')['Open']
and a month end date to show how relativedelta treats this case:
today = date(2016, 1, 31)
Create the sequence of dates:
n_months = 30
dates = [today - relativedelta(years=m // 12, months=m % 12) for m in range(n_months)]
to get:
stock_data.loc[dates]
Date
2016-01-31 108.989998
2015-12-31 106.000000
2015-11-30 105.839996
2015-10-31 104.510002
2015-09-30 88.440002
2015-08-31 90.599998
2015-07-31 94.949997
2015-06-30 86.599998
2015-05-31 79.949997
2015-04-30 80.010002
2015-03-31 82.900002
2015-02-28 80.680000
2015-01-31 78.000000
2014-12-31 79.540001
2014-11-30 77.669998
2014-10-31 74.930000
2014-09-30 79.349998
2014-08-31 74.300003
2014-07-31 74.000000
2014-06-30 67.459999
2014-05-31 63.950001
2014-04-30 57.580002
2014-03-31 60.779999
2014-02-28 69.470001
2014-01-31 60.470001
2013-12-31 54.119999
2013-11-30 46.750000
2013-10-31 47.160000
2013-09-30 50.139999
2013-08-31 42.020000
Name: Open, dtype: float64